Some servants always tell the truth whereas other servants always lie

Question

Suppose in a house there are exactly two types of servants: those who always speak the truth and those who always lie. A visitor arrives at this house to meet the owner and one of the servants says "My master is in the house if and only if I am telling the truth". What should the visitor do in order to reveal the truth?

Is the problem not well posed to arrive at any solution?

It's natural to assume that the visitor is aware of these only two categories of servants, but does not actually know the category of the teller. I am completely stuck in the half part of my approach to the solution. If this servant is of first category (truth saying) then the master is surely in house. But what if it is the other way round? Please help.

Answer 1

You don't need to ask any questions! The master is home!

If the master were out, a truth telling servant couldn't say that. That's obvious.

If the master were out and the servant were a liar then "If I am a truth teller then X" would be true. And "If the master is in then Y" would be true. So "I am a truth teller if and only if the master is in" would be !!!true!!!.

So that statement can only be said if the master is home.

Truth value of "I am truth teller If and only if X": 1)$T \iff T$ is True.

2)$T \iff F$ is false.

3)$F \iff T$ is false.

4)$F \iff F$ is true.

A truth teller can only do 1 or 4 but not 4. A liar can only do 2 or 3 but 2. So only 1 or 3 is possible. In both cases, X is true.

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More clarification: Suppose a liar.

Liar: "X". We can conclude "not X".

Liar: "Something true and X". If X is true the whole thing is true, so X is false.

Liar: "If Something false then X". As something false isn't true anything can result so the statement is true no matter what X is. The universe just imploded.

Liar: "If Something false then X and Y". As "If something false then X" is true, we must conclude not Y.

Liar: "If I am telling the truth then X and Y." "I am telling the truth" $\subset$ "something false". We conclude not Y.

Liar: "If I am telling the truth then X and If A then B". We must conclude "not(If A then B)". If A is false then anything that follows is true. So A is true. And so to be false, B must be false. We must conclude A and not B.

Liar: "If I am telling the truth then X and If A then I am telling the truth". This is fine as "I am telling the truth" is false. So we conclude X can but either true or false and A is true.

Liar: "If I am telling the truth then (the master is home) and If (the master is home) then I am telling the truth". We must conclude the master is home by the second clause. The first clause is inclusive. So the master is home.

Liar: "I am telling the truth if and only if the master is home". Same as above. The master is home.

Answer 2

All the visitor has to do is think. The visitor can already deduce that the master is in the house. If the servant is a truth teller, then the master is in the house by the if-and-only-if clause. If the servant is a liar, then it is not true that he is telling the truth, and therefore the right side of the if-and-only if statement is not satisfied. This would imply that the master is not in the house if it were not for the fact that the if-and-only-if statement is false since the servant is a liar. Therefore the master is in the house in this case as well.

Answer 3

If being in house is $h$ and telling truth is $T$

$$h = T$$

What was told to the visitor is the above math. One can lie about it in a number of ways (i.e. the reality could be any of the following):

$$h = 0; \text{ The master is not at home and it has no dependence on telling truth}$$ $$h = 1; \text{ The master is at home and it has no dependence on telling truth}$$ $$h = \bar{T}; \text{ Exact opposite of what was told}$$

So if the servant is telling truth, then master is at home. If the servant is lying, it is not conclusive as to which of the truths listed above is the one being lied about. Actually the third one results in $h=1$ because $T=0$ for lies.